Question

$K_P=0.04$ atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of $C_2{H}_6$ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
$C_2{H}_6\left(g\right)\rightleftharpoons C_2{H}_4\left(g\right)+H_2\left(g\right)$

• A. 7.24 atm
• B. 3.62 atm
• C. 1 atm
• D. 1.5 atm

Answer 1

The correct option is B 3.62 atm

$\begin{array}{cccccc}& C_2{H}_6\left(g\right)& \rightleftharpoons & C_2{H}_4\left(g\right)& +& H_2\left(g\right)\\ t=0& 4atm& & 0& & 0\\ & -x& & x& & x\end{array}$
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$\begin{array}{cccccccccccc}t=t_{eq}& 4-x& & & & x& & & & & & x\end{array}$

$0.04=\frac{x^2}{4-x}$
On solving, $x=0.38;$
$P_{C_2{H}_6}=3.62atm$